To solve the questions provided, let’s go step by step.
1. Proving that \( S = -(100 + m)X + \frac{L}{m} \)
Given that \( S \) and \( X \) are dependent variables and the problem seems to involve a linear relationship, you can assume a linear model of the form: \[ S = aX + b \]
Where \( a \) is the slope, and \( b \) is the y-intercept. In your case, \(a\) becomes \(-(100 + m)\), and \(b\) becomes \(\frac{L}{m}\), as stated in the final form of the equation.
To prove this, we need additional details about the variables \(S\), \(X\), \(L\), and their relationship to \(m\). Without a more detailed context, we can't fully derive the expression, but if we're working through a physics context involving torque or gravity, it might be assumed that the equation reflects an equilibrium situation or the balance of moments in a system.
2. Plotting a graph of \( S \) against \( X \)
Create a table of values based on the two sets you have and calculate their corresponding \(S\) values with error included. The values should be plugged into the derived equation or directly plotted from the provided data.
Here’s the data you have:
| \( S \) (cm) | \( X \) (cm) | |---------------|---------------| | 2.5 ± 0.05 | 21.6 ± 0.05 | | 5.0 ± 0.05 | 20.6 ± 0.05 | | 10.0 ± 0.05 | 18.1 ± 0.05 | | 20.0 ± 0.05 | 13.9 ± 0.05 | | 30.0 ± 0.05 | 9.3 ± 0.05 | | 40.0 ± 0.05 | 50 ± 0.05 |
Using this data, the graph should plot \(S\) on the vertical axis (Y-axis) and \(X\) on the horizontal axis (X-axis).
3. Finding the slope
To find the slope of the line once you have the graph plotted, you can use the slope formula:
\[ \text{slope} = \frac{\Delta S}{\Delta X} \]
Choose two points from your plotted line, \((S_1, X_1)\) and \((S_2, X_2)\), and substitute them into the formula:
\[ \text{slope} = \frac{S_2 - S_1}{X_2 - X_1} \]
You might choose endpoints or any two points on the line.
4. Finding the value of \(S\) for \(X = 0\)
From the graph, find where the line intersects the Y-axis. This point corresponds to the value of \(S\) when \(X=0\). You can also calculate \(S\) using the derived equation \(S = -(100 + m) \cdot 0 + \frac{L}{m}\), which simplifies to \(S = \frac{L}{m}\).
5. Deducing the value of \(m\) from the graph
If you have already determined the slope (let's say it is \(m_s\)) and found the intercept \(b\) from the graph (which is \(\frac{L}{m}\)), then the relationship would yield:
- From the linear equation \(S = -(100 + m)X + \frac{L}{m}\), you equate the slope \(- (100 + m)\) to the slope found from the graph.
- Knowing \(b = \frac{L}{m}\), you can rearrange this to find \(m\) through substitution.
Assume you know the value of \(L\) (needed for calculating \(m\)), use it appropriately with your slope and y-intercept findings to conclude \(m\).
Conclusion
If you provide more specific contexts or numerical output from the graph, we can delve deeper into actual calculations for the slope or other values. With the general formulas and methodologies outlined above, you should be able to work through the specifics in any calculation software or using graphing techniques.